Question: What is the inverse of the function $f(x)=\dfrac{3+4x}{1-5x}$ ? $ f^{-1}(x) =$
Solution: Let's start by replacing $f(x)$ with $y$. $y=\dfrac{3+4x}{1-5x}$ Now let's swap $x$ and $y$ and solve for $y$. $\dfrac{3+4y}{1-5y}=x$ [Why do we swap x and y?] $\begin{aligned} \dfrac{3+4y}{1-5y}&=x \\\\ 3+4y&=x(1-5y) \\\\ 3+4y&=x-5xy \\\\ 4y+5xy&=x-3 \\\\ y(4+5x)&=x-3 \\\\ y&=\dfrac{x-3}{4+5x} \end{aligned}$ In conclusion, this is the inverse function: $f^{-1}(x)=\dfrac{x-3}{4+5x}$ [I saw someone solve this problem by originally solving for x. Were they wrong?]